Sea water contains 29.25 % NaCl and 19 % MgCl₂ by weight of solution. The normal boiling point of the sea water is _____________ ° C (Nearest integer)
Assume 100 % ionization for both NaCl and MgCl₂
Given : K_b(H₂ O)=0.52 K kg mol⁻¹
Molar mass of NaCl and MgCl₂ is 58.5 and 95 g mol⁻¹ respectively.

Solution

Chemistry ma’am · 4 steps

Take 100 g of sea water, count every ion, and divide by the water alone, not the whole solution.

1. Moles in 100 g. In 100 g of solution there is 29.25 g NaCl, which is 29.25/58.5 = 0.5 mol. There is also 19 g MgCl₂, which is 19/95 = 0.2 mol.
2. Mass of solvent. Molality needs the mass of the solvent, so subtract both salts. Water = 100 - 29.25 - 19 = 51.75 g, which is 0.05175 kg.
3. Count the particles. With full ionization, NaCl gives 2 ions (i = 2) and MgCl₂ gives 3 ions (i = 3). Total moles of particles = 0.5 × 2 + 0.2 × 3 = 1.6 mol.

   Δ T_b = K_b × (Σ i n)/W_(water) = 0.52 × 1.6/0.05175 = 0.52 × 30.92 = 16.08 K

4. So. The boiling point is 100 + 16.08 = 116.08 °C, which rounds to 116.

The common slip: 108. Taking 100 g as the mass of water instead of 51.75 g gives Δ T_b = 8.32 K and a boiling point of 108, not 116.

The idea to keep. When a question gives a percentage by weight of solution, assume 100 g of solution and subtract the solute to get the solvent mass. Then add up i × n for every solute before you use K_b.

Answer from the NTA final answer key. Solution by PrepWiser.